Cayley graphs formed by conjugate generating sets of Sn 1
Some Terminology
In this paper, we will let N denote the set {1, 2, . . . , n}. Sn will denote the symmetric group acting on n elements with canonical action on N . An will denote the alternating group acting on N . We will use the notation (a1 a2 . . . ak1 )(b1 b2 . . . bk2 ) . . . to express a permutation as a product of disjoint cycles. By the support of a permutation we will mean those elements not fixed by the permutation. Given a tuple A = (a1 , . . . , ak ), we define its extended conjugacy class in Sn to be the set of all permutations such that, when decomposed into disjoint cycles, contain cycles of lengths a1 , . . . , ak , and no others (the tuple A may contain duplicate elements). We denote it by C(A). Given a set S ⊂ N , define the subsymmetric group of S as the set of all permutations in Sn that fix all elements outside of S. Define the subalternating group of S as the set of all even permutations in Sn (i.e., the permutations in An ) that fix all elements outside of S. A semisymmetric group of S is defined as a subgroup of Sn that stabilizes S whose restriction to S forms a symmetric group acting on S. A semialternating group of S is defined similarly. Given a graph Γ, we let V (Γ) and E(Γ) denote the vertices and edges of Γ, respectively. An Eulerian path is a walk in Γ that traverses each edge exactly once. It is called an Eulerian cycle if the first and last vertices in the walk are the same. Given a group G and a set S ⊂ G, the Cayley graph Γ = Cay(G, S) is defined as follows: each vertex is an element of G, and two vertices g, h ∈ V (Γ) are adjacent if gh−1 ∈ S or hg −1 ∈ S.
2
Motivation and Overview
Cayley graphs are of general interest in the field of Algebraic Graph Theory and also have certain properties desirable in practical applications. We present here a brief survey of some of the broader results and conjectures surrounding Cayley graphs. Godsil and Royle [6] provide a useful overview of work on graphs with transitive permutation groups in general, which we partially reproduce here. First, all Cayley graphs are vertex-transitive since the mapping φg (x) = xg is an automorphism for 1
all g ∈ G. As such, there is always a representation of G in Aut(Cay(G, S)), denoted R(G). R(G) acts not only transitively but regularly on the vertices of Cay(G, S). Sabidussi has shown that the converse of this is true, namely that Γ if a Cayley graph of G if and only if Aut(Γ) contains a subgroup isomorphic to G that acts regularly on V (Γ). We also have that every vertex-transitive graph is a retract1 of a Cayley graph, a theorem akin to the famous result that every finite group is a subgroup of the symmetric group. There are also theorems showing that vertex-transitive graphs in general have high vertex and edge connectivity, which points to uses in practical applications. Specifically, Cayley graphs have been used to create networks with small diameter and valency and high connectivity for uses in parallel processing, and Schreier coset graphs, a generalization of Cayley graphs, have been used to solve certain routing problems [2]. See also [1] and [13]. One of the most tempting conjectures related to Cayley graphs is known as the Lovasz conjecture, stating that every Cayley graph has a Hamiltonian cycle. For more information, see [3], [8], and [11]. While we offer some computational ideas in relation to the Lovasz conjecture, the focus of this paper is on another difficult problem in Algebraic Graph Theory, that of characterizing the automorphism groups of Cayley graphs. We have a poor understanding of the automorphism groups of Cayley graphs, though there are some notable exceptions (see below). These groups are fundamental as the most natural algebraic structure to associate with an arbitrary highly symmetric graph. Note that a graph can be defined as a collection of vertices and edges. Two vertices are adjacent if there exists an edge connecting them, and two vertices v1 and v2 are connected if there exists a sequence of adjacent vertices containing v1 and v2 . On the other hand, consider the following definition: Given a collection of vertices V and a collection of edges E, we can let each element of E act on V as a transposition swapping the two vertices on which E is incident. If we then let multiplication in E extend through the definitions of a group action, E generates a subgroup of the symmetric group acting on V (we denote this subgroup as < E >). Then we say that v1 and v2 are adjacent if (v1 v2 ) ∈ E, and that v1 and v2 are connected if (v1 v2 ) ∈< E >. Additionally, connected components correspond to orbits of V under E. A tree is a minimal generating set of S|V | consisting 1
A retract of a graph Γ is a subgraph Γ0 such that there exists a homomorphism φ of Γ onto Γ0 whose restriction to Γ0 is the identity.
2
only of transpositions (thus the fact that trees have n − 1 vertices corresponds to the fact that it takes n − 1 transpositions to generate Sn ). It is easily verified that these definitions are equivalent. The algebraic properties of the related Cayley graphs of trees in the above definition are wellunderstood. We know in particular that the Lovasz conjecture holds for these graphs. Furthermore, in 2003 Feng [4] generalized a result by Godsil [6] that fully characterizes the automorphism groups of these graphs. The above definition of a graph in terms of transpositions can be generalized. Given a collection of vertices (which, from now on, for convenience, will without loss of generality be N ), and a set T ⊂ Sn in which all elements of T are conjugate (say with conjugacy class C), then we can define elementary notions in a C-graph as follows. v1 , v2 ∈ N are adjacent if they have the same orbit under a single element of T . They are semi-connected if they have the same orbit under T , and connected if (v1 v2 ) ∈< T > (it is then easy to verify that semi-connectivity and connectivity are equivalence relations). Connected components correspond to subsymmetric groups of < T >. A tree is a minimal generating set of Sn with all elements lying in C. It is natural to ask why we add the somewhat artificial-looking stipulation that all elements of T belong to the same conjugacy class. The main reason is that this stipulation is inherent in the construction of a normal graph, where all edges are transpositions. Additionally, without this restriction we get the result that a tree, under our fairly intuitive definition, almost always has 2 edges since (1 2) and (2 3 4 . . . n) generate Sn . In this paper we will characterize C-trees and study some of their properties, including a generalization of Feng’s result. However, we will still use the language of graphs for the sake of intuition. For approaches to extending the above intuitive generalization to a well-structured system, see the concluding section on open problems.
3 3.1
Trees Additional Definitions for Trees
A set T ⊂ Sn is said to be semi-connected if N has a single orbit under T (i.e. all elements of N are semi-connected). We call it split if the intersection of the supports of any two elements of T has size
3
at most one. Note that if T generates Sn then it must be connected. Given a tuple A and integer n, we define f (A, n) to be the infimum of |G| across all G ⊂ C(A) that generate Sn . We aim to find f (A, n) for every A for sufficiently large n. Let c(A) be defined as |A| X
ai − 1
i=1
We aim to prove that there exists a function X0 (A) such that, for n ≥ X0 (A), f (A, n) is equal to
d
n−1 e c(A)
(1)
when c(A) is odd, and ∞ otherwise. When c(A) is odd, then A defines the conjugacy class of an even permutation and so f (A, n) is obviously ∞ (because it is impossible to generate any odd permutations). Note further that f (A, n) is necessarily at least (1), as c(A) counts the number of transpositions necessary to generate an element of C(A), and so if it was smaller then it would be possible to generate Sn with less than n − 1 transpositions. Another way to see this is that no potential generating sets can be semi-connected, and thus cannot generate Sn .
3.2
Some Motivation: Cycles
We study first the case of a single k-cycle, i.e. |A| = 1 and a1 = k. We will give explicit generators for S2k−1 : Proposition 3.1 The set {(1 2 . . . k), (k k + 1 . . . 2k − 1)} generates S2k−1 . Proof We construct something similar to a semisymmetric group of {1, 2, . . . , k}, except with the elements lying in the positions {k, k + 1, . . . , 2k − 1}. From this we will construct a subsymmetric group of {1, 2, . . . , k}, which will finally allow us to construct the entire symmetric group S2k−1 . Lemma 3.2 We can place the elements {1, 2, . . . , k} in any order in positions {k, k + 1, . . . , 2k − 1} as long as we allow the other elements to move arbitrarily. Proof Let σ = (1, 2, . . . , k) and τ = (k, k + 1, . . . , 2k − 1). If we allow the first k − 1 elements to be arbitrary, we can set the last k elements to any permutation π = {π1 , π2 , . . . , πk } of {1, 2, . . . , k} as 4
follows: Rotate π1 to the kth position using σ, then apply τ once. Now rotate the second element in the permutation to the kth position (again with σ), and apply τ again. Continue this process until we have put all k of the desired elements into place. For example, to make the last 4 elements {3, 1, 4, 2}, we would apply σ 1 τ σ 2 τ σ 1 τ σ 2 τ . Each successive set of applications of σ moves the next desired element into place (3, then 1, then 4, then 2). Lemma 3.3 For k even, the k-cycles generate Sn for n ≥ k. Moreover, for k odd, the k-cycles generate An . Proof Let U be the set of all k-cycles. U is closed under conjugation, so < U > is normal in Sn . Thus < U > is either (e), D4 , An , or Sn , where by D4 we mean the normal subgroup of A4 isomorphic to the dihedral group of order 4. It can’t be (e) because U is non-trivial, and it can’t be D4 because D4 contains no cycles. Thus it is An or Sn . If k is odd, it must be An because U consists of only even permutations. If k is even, it must be Sn since U contains an odd permutation. This completes the lemma. Lemma 3.4 We can generate the subsymmetric group on {1, 2, . . . , k}. Proof Take some such permutation π generated in the manner of Lemma 3.2, and consider πτ π −1 . This creates an arbitrary k-cycle among the first k elements while fixing the last k − 1 elements. Then by Lemma 3.3 we can generate the subsymmetric group on {1, 2, . . . , k}. Our Lemma is thus complete. Now, to generate an arbitrary permutation π = {π1 , . . . , π2k−1 } in S2k−1 , first use τ to move {πk+1 , . . . , π2k−1 } to the first k − 1 elements of the set. We can do this by moving each one to the kth position, then, since we can generate any permutation among the first k elements (recall that we can do this by Lemma 3.4), move it to an arbitrary place among the first k − 1 elements in which we haven’t already put anything with this process. Next apply a permutation that puts {πk+1 , . . . , π2k−1 } in the proper order (though leaving them in the positions {1, 2, . . . , k − 1}). We can then move them to their proper locations with (στ )k−1 . Now that the last k − 1 elements are in place, we can apply whatever permutation is necessary to put the first k elements in place. We can thus generate an arbitrary permutation and therefore S2k−1 . This completes Proposition 3.1. 5
Proposition 3.5 f ((k), n(k − 1) + 1) = n for n ≥ 2. Proof This follows by induction on n. Proposition 3.1 proves the base case of n = 2. The inductive step is completed by the following easily verified lemma: Lemma 3.6 The subsymmetric group on S, together with the cycle (a1 , . . . , an ), generates the subsymmetric group on S ∪ {a1 , . . . , an } provided that S ∩ {a1 , . . . , an } = 6 ∅ and S 6⊂ {a1 , . . . , an }. Corollary 3.7 f ((k), n) = d n−1 e for n ≥ 2k − 1. k−1 Proof Take the construction for when
n−1 k−1
is an integer (i.e. that given above in Proposition 3.5).
Then, to extend the formula to arbitrary n, add the k-cycle (n − k + 1, n − k + 2, . . . , n) and apply Lemma 3.6. This completes the claimed characterization of f (A, n) for cycles.
3.3
More Motivation: Products of Transpositions
Having proven our result for cycles, we would like to extend it to more complex permutations. We will start with the simplest of these, i.e. products of disjoint transpositions. We call a permutation basic if it is of this form, and denote Bk = (2, 2, . . . , 2) (k twos). Proposition 3.8 f (Bk , n) = d n−1 e for n ≥ k(2k + 1) + 1. k Proof Our goal will be to write 2k + 1 explicit generators for Sk(2k+1)+1 . Then we can easily proceed by induction as before. Such generators must form a semi-connected set. However, we would also like the set to be split so that only one cycle interacts at a time when multiplying permutations. To do this, we will find an Eulerian cycle of K2k+1 , which will allow us to create a connected and split set. First note that every vertex of K2n+1 has even degree (in particular, degree 2n), so that K2n+1 has an Eulerian cycle. We now construct generators from this cycle. We start with an example. Consider k = 3, 2k + 1 = 7, and the cycle 1 → 2 → 4 → 7 → 1 → 3 → 6 → 7 → 2 → 5 → 6 → 1 → 4 → 5 → 7 → 3 → 4 → 6 → 2 → 3 → 5 → 1. We have the generators
g1 = (1 2)(5 6)(12 13) g2 = (2 3)(9 10)(19 20) g3 = (6 7)(16 17)(20 21) g4 = (3 4)(13 14)(17 18) 6
g5 = (10 11)(14 15)(21 22) g6 = (7 8)(11 12)(18 19) g7 = (4 5)(8 9)(15 16)
Note that if we follow the path of permutations containing (1, 2), (2, 3), (3, 4), . . . , (21, 22), then we get g1 , g2 , g4 , g7 , . . . , g3 , g5 , i.e. exactly the constructed Eulerian cycle (with the exception of the final vertex). This is the general method in which we will construct our generators. Note that the properties of an Eulerian cycle guarantee that these generators will be semi-connected and split. The semi-connected part is obvious, whereas the split part is a consequence of the fact that every edge is traversed exactly once, which corresponds to the fact that each pair of generators move at most one common element. We now show this construction generates Sk(2k+1)+1 . Theorem 3.9 If T ⊂ C(Bk ) is semi-connected and split, then < T >= Sn or An , depending on whether k is even or odd. Proof We call two generators adjacent if they move a common element. Consider two adjacent generators, gi and gj , and consider gi gj gi gj . All transpositions are applied twice in this case and therefore cancel, except for the two transpositions that act on the same element, which multiply to a three cycle. So, in the above example, g5 g7 g5 g7 = (14 15)(15 16)(14 15)(15 16) = (14 16 15)2 = (14 15 16). In this manner, we generate all 3-cycles of the form (i i + 1 i + 2). We wish to show that these generate An . From this, we would be done, since any odd permutation then allows us to generate Sn . In fact, it is convenient for later purposes to prove a slightly stronger result: Lemma 3.10 The subalternating group on S, together with the cycle (a1 , . . . , an ), generates the subalternating group on S∪{a1 , . . . , an } provided that S∩{a1 , . . . , an } = 6 ∅ and that |S∩{a1 , . . . , an }| ≤ |S| − 2. In addition, if n is even, it generates the entire subsymmetric group. Proof Like Lemma 3.6, the proof is easy enough to omit. The only important detail to note is that we get
|S∩{a1 ,...,an }|! 2
distinct permutations, which must be the alternating group when n is odd or must
generate the symmetric group by Lagrange’s theorem when n is even. In particular, a 3-cycle looks like A3 , so the given 3-cycles indeed generate the alternating group (they are semi-connected since T was semi-connected), and we are done with Theorem 3.9. 7
Our proof of the remainder of Proposition 3.8 (i.e. the induction and extension to cases when c(A) does not divide the n − 1) follows in exactly the same manner as that of Proposition 3.1, and so we omit it, instead referring the reader to Proposition 3.5 and Corollary 3.7. The only necessary modification is that we must deal with each of the cycles in the added permutation one at a time in our inductive step.
3.4
The Main Proof
We would next like a general criterion for connectedness. We present it here: Definition A set T ⊂ C(A) is called balanced if it is possible to divide the set of orbits of elements of T into disjoint sets S1 , S2 , . . . , S| A| such that all orbits in Si have the same size and each element of Si overlaps with at least one other element of Si . We will denote the size of the orbits in Si by ai . Theorem 3.11 All semi-connected, split, balanced sets generate Sn . Proof We proceed by induction on two quantities: first |A|, then the number of occurrences of 2 in A. Note that we have already proven the base cases of this induction in Corollary 3.7 and Proposition 3.8. We call two permutations i-adjacent if they both have orbits in Si . Pick i such that ai is maximal, and consider any two i-adjacent permutations, σ and τ , with orbits i σ and i τ in Si . By the same argument as Proposition 3.1, these generate the semialternating group on the elements moved by the two identified cycles (moving 2ai −1 elements in total). Thus in particular, by Chebyshev’s Theorem [12], there exists a prime strictly between ai and 2ai , and the semialternating group contains a cycle of this length, call it p. Consider each cycle of this length in our semialternating group, and apply it lcma∈A a times. Since p is prime and aj < p for each j, we end up with a p-cycle, which we can then apply some number of times to get back to our original p-cycle. Note, however, that all other elements that were moved contained only cycles of length aj for some j, and so were all cancelled out by the above repeated application. Thus we are left only with the actual p-cycle. Repeating this for all such p-cycles in the semialternating group gives us all actual p-cycles, i.e. those living in the associated subalternating group. Thus, by the same arguments as in Lemma 3.3, they generate the entire subalternating group. 8
If ai is odd, then i σ and i τ both live in this subalternating group, and so we can take σ(i σ)−1 and τ (i τ )−1 . Taking σ(i σ)−1 for all σ ∈ T (we can do this since T is balanced) gives us a semi-connected, split, balanced set with strictly less orbits in each permutation, so that we can apply the inductive step to generate the subsymmetric group on the elements moved by these new permutations. Then, by adding i σ for each σ ∈ T , by Lemma 3.5 we can generate the entire symmetric group, and we are done. On the other hand, if ai is even, then we can only cancel i σ down to a transposition. However, this gives us an extended conjugacy class with the same number of orbits, but with strictly more occurrences of 2 in A than before. Thus we can apply the inductive step in the same manner as above, and are once again done. We are now ready to prove our major contention: n−1 Theorem 3.12 Let A be a k-tuple. Then there exists some X0 (A) such that f (A, n) = d c(A) e when n ≥ X0 . Furthermore, X0 ((k)) ≤ 2k − 1, X0 (Bk ) ≤ 2k+1 + 1, and X0 (A) ≤ c(A)Φ2|A| (2|A|) + 1, 2
where Φk denotes the kth cyclotomic polynomial. Proof Note that the first two bounds have already been proven. For the final case, we again use Eulerian cycles, this time with the goal of creating a semi-connected, split, and balanced set. In particular, we find a prime congruent to 1 mod 2k. We know that such a prime exists that is less than Φ2k (2k).2 Take such a prime, p, fitting the properties described above. If
p−1 2
= kn, then we will work in the
extended conjugacy class that is equivalent to n copies of A, then use this to move down to A itself. We construct an Eulerian cycle for Kp as follows. The edges (mod p) will be
1, 2, . . . , p, 2, 4, . . . , 2p, 3, . . . , 3p, . . . ,
p−1 p(p − 1) , p − 1, . . . , 2 2
So, for example, if p = 7 then we have (in the case of 2-cycles) the associated generators
(1 2)(11 12)(19 20) (2 3)(8 9)(17 18) 2
Since this result is not well-known, we offer a proof sketch in the Appendix.
9
(3 4)(12 13)(15 16) (4 5)(9 10)(20 21) (5 6)(13 14)(18 19) (6 7)(10 11)(16 17) (7 8)(14 15)(21 22)
We can extend this past 2-cycles (for example, permutations in the extended conjugacy class (2, 4, 5) in the following manner:
(1 2)(11 23 24 12)(19 37 38 39 20) (2 3)(8 25 26 9)(17 40 41 42 18) (3 4)(12 27 28 13)(15 43 44 45 16) (4 5)(9 29 30 10)(20 46 47 48 21) (5 6)(13 31 32 14)(18 49 50 51 19) (6 7)(10 33 34 11)(16 52 53 54 17) (7 8)(14 35 36 15)(21 55 56 57 22)
Note that we simply add elements to cycles in the ith column until the cycles in that column have length ai . Note also that this is a balanced set by construction. It is easy to verify that this also defines an Eulerian cycle, and is thus connected and split. On the other hand, we have the following result: Lemma 3.13 If B is equivalent to k copies of A, and if there exists T ⊂ C(B) that generates Sn , then there exists T 0 ⊂ C(A) that generates Sn , and furthermore such that |T 0 | = k|T |. Proof Split each σ ∈ T into k permutations such that each of these permutations lies in C(A). These obviously generate Sn since products of them generate Sn . n−1 This proves the base case of a final induction showing that f (A, n) = d c(A) e for all n ≥ X0 ,
where X0 = pc(A) + 1. This induction will finally prove Theorem 3.12. However, once again this new induction is identical to the completions of Propositions 3.1 and 3.8, and so we refer the readers there for the completion of the proof.
4
Automorphism Groups
We devote this section to the characterization of the automorphism groups of certain C-graphs. Definition Given a split set of cycles T ⊂ Sn , the cycle graph Cyc(T ) is formed by associating each vertex with an element of N and drawing an edge between vertices i and j if there exists an element mapping i to j. When T consists of transpositions, Feng [4] refers to Cyc(T ) as T ra(T ). 10
Definition Given a split set of cycles T , the degree of some t ∈ T is defined as the number of distinct points in its support that overlap with other cycles. If t has degree 1, we call it a leaf. Definition A split set of cycles generating Sn is said to be normal if any element is adjacent to at most 1 leaf, and furthermore Cyc(T ) is a tree (note that this is stronger than asking that T be a minimal generating set of Sn , as it effectively adds the criterion that n ≡ 1 (mod k), where T consists of k-cycles). We use this to offer a partial generalization to a theorem by Feng [4] that states that Aut(Cay(Sn , T )) ∼ = R(Sn ) o Aut(Sn , T ), where Aut(Sn , T ) = {φ ∈ Aut(Sn ) | φ(T ) = T }, and furthermore that Aut(Sn , T ) ∼ = Aut(T ra(T )). In the following, T will always be normal, and if we talk about a graph it will be Cay(Sn , T ) unless otherwise specified: Theorem 4.1 Let T be a normal set. Then Aut(Cay(Sn , T )) ∼ = R(Sn ) o Aut(Sn , T ), where R(Sn ) is the representation of Sn as an action on Cay(Sn , T ). Proof We use Feng’s idea of finding cycles that force graph automorphisms to be group automorphisms. Certain lemmas requiring case analysis will be dealt with in the appendix. Lemma 4.2 Let t1 , t2 ∈ T . Then there exists a unique 4-cycle containing the path t2 → (e) → t1 iff t1 t2 = t2 t1 , and furthermore the cycle will be (e) → t1 → t1 t2 → t2 → (e). Proof See appendix. Lemma 4.3 Let t1 , t2 ∈ T such that t1 t2 6= t2 t1 . Then the 6-cycle corresponding to t1 t2 t1 t2 t1 t2 is sent to another cycle of this form under graph automorphisms when t1 and t2 are transpositions. −1 −1 −1 −1 −1 Otherwise, the same statement holds for the 12-cycle corresponding to t1 t2 t−1 1 t2 t1 t2 t1 t2 t1 t2 t1 t2 .
Proof The case of transpositions was dealt with by Feng [4]. It is easily verified that the latter construction is a cycle when t1 and t2 are not transpositions (it is the union of two cycles when they are transpositions). Also note that no two consecutive edges correspond to commuting generators, and this property is preserved through graph automorphisms by Lemma 4.2. It is natural to try to prove that this is the only 12-cycle going through t1 and t2 where no two consecutive edges commute. However, this 11
is false, as shown by the following counterexample: Let a = (1 2 3 4), b = (1 5 6 7), c = (1 8 9 10), d = (1 11 12 13). Then aba−1 b−1 aba−1 b−1 aba−1 b−1 = abcdcb−1 a−1 bc−1 d−1 c−1 b−1 = (e). However, if we only allow use of the symbols a, b, a−1 , b−1 , then this is indeed the only noncommuting 12-cycle, as demonstrated in the appendix. This leads to a proof of our theorem in a special case, which we will make use of: Lemma 4.4 Theorem 4.1 holds when |T | = 2. Proof The preceding comments show us that commutators of generators map to commutators of generators. Thus φ(a)φ(b) = φ(ab) for all generators a, b, so that φ(x)φ(y) = φ(xy) for all xy by induction. The induction itself is sufficiently non-trivial that we feel obliged to include it, but sufficiently technical that we will relegate it to the appendix, even though it requires no case analysis. We have thus shown that all graph automorphisms fixing (e) are in fact group automorphisms as well. That this implies Theorem 4.1 we wait to prove in full generality at the end of this section. Now for any a, b ∈ T , look at Γ0 = Cay(Sn , {a, b}) ⊂ Γ. The 12-cycle described above must lie inside Γ0 . We wish to show that, for any automorphism φ ∈ Aut(Γ) fixing (e), φ(Γ0 ) = Cay(Sn , {φ(a), φ(b)}), from which it will follow that commutators map to commutators in general, and we will have proved Lemma 4.3, whence Theorem 4.1 follows from the same arguments as in Lemma 4.4. We will, in fact, prove a stronger contention, namely that if two edges represent the same group element, then their edges also represent the same group element. We first offer an automorphisminvariant criterion for determining whether two adjacent edges represent the same group element of the Cayley graph when T is normal. Lemma 4.5 Let x → y → z be a path in Γ. Then xy and yz represent the same group element if and only if the number of 4-cycles going through xy equals the number of 4-cycles going through yz. Proof Note that if xy and yz correspond to the same group element, then the number of 4-cycles going through xy definitely equals the number of 4-cycles going through yz by Lemma 4.2. (Note that this is true even if T consists of 4-cycles.) The opposite direction is an easy consequence of the normality condition. 12
Now note that, by looking at commutativity of edges, we obtain the incidence structure of Cyc(T ). Thus the group elements that each edge corresponds to is uniquely determined by which edge each leaf corresponds to (this is simply a consequence of the fact that a tree is determined by the paths between terminal nodes). Thus, given an edge from v corresponding to a leaf λ, whose pre-image under φ is λ0 , it suffices to prove that any edge from an adjacent vertex w corresponding to λ also has pre-image λ0 . First note that unless |T | = 2, which has already been dispatched of, all leaves commute. We consider two cases: adjacency between v and w is induced by a leaf, or the adjacency is induced by a non-leaf. Case one: We may assume that all leaves commute, whence we are done by Lemma 4.2. Case two: By the normality condition, the group element associated with vw must commute with all but one edge, from which we are again done by Lemma 4.2. Then, noting that leaves are mapped to leaves under any graph automorphism, the final leaf only has one place to go (actually, one could make the argument that there are two places to go – to itself or to its inverse, but both of these edges correspond to the same group element, which is all that we care about). This completes our contention, so that we are finally done with Lemma 4.3. By Lemma 4.2, commutativity of edges is preserved through graph automorphisms. Furthermore, cycles are preserved through graph automorphisms. Thus in particular, {φ(t1 ), φ(t1 )φ(t2 ), φ(t2 ), (e)} must be the image of {t1 , t1 t2 , t2 , (e)} if φ is a graph automorphism fixing (e) and t1 , t2 ∈ T commute. This implies that φ(t1 )φ(t2 ) = φ(t1 t2 ). By the same argument, and using Lemma 4.3, φ(t1 )φ(t2 ) = φ(t1 t2 ) if t1 , t2 ∈ T don’t commute. Thus φ(t1 )φ(t2 ) = φ(t1 t2 ) for all t1 , t2 ∈ T . This implies that φ is not only a graph automorphism but a group automorphism, by the same argument as in Lemma 4.4. It follows by abuse of notation that Aut(Cay(Sn , T ))(e) ⊂ Aut(Sn , T ), where Gx denotes the stabilizer of x under the action of G. But it is well-known that Aut(Sn ) ∼ = Sn (the isomorphism being with the inner automorphism group) for n 6= 6 [4], so that Aut(Sn , T ) ⊂ Aut(Cay(Sn , T ))(e) when n 6= 6 (it is easily verified that any inner automorphism of Sn preserving T must also preserve incidence in Γ and is thus a graph automorphism). Note that n = 6 only when k = 2, which has already been dispatched, so the theorem holds for all n that we care about. Since Aut(Cay(Sn , T )) = R(Sn )Aut(Cay(Sn , T ))(e) and the two subgroups have trivial intersection, we will have a complete
13
characterization of Aut(Cay(Sn , T )) if we can show that R(Sn ) is normal. This follows since R(Sn ) is closed under conjugation by elements of Aut(Sn , T ). Thus we have that Aut(Cay(Sn , T )) ∼ = R(Sn ) o Aut(Sn , T ), as stated. Comment Though it is always regrettable when a result cannot be proven in full generality, we claim that the normality condition is relatively weak. Indeed, given any set T , we can define a normalization of T to be a new C-graph obtained from T by adding another cycle incident on each leaf of T . It is easily verified that this results in a normal set. We end this section with a simple lemma: Lemma 4.6 Let T = {(1 2 . . . k), (1 k + 1 . . . 2k − 1), . . . , (1 n(k − 1) + 2 . . . n(k − 1) + k)}. Then Cay(Sn(k−1)+k , T ) is edge-transitive. Proof As it is obviously vertex-transitive, it suffices to observe that (2 k + 1 . . . n(k − 1) + 2)(3 k + 2 . . . n(k − 1) + 3) . . . (k 2k − 1 . . . n(k − 1) + k) performs a cyclic shift on the elements of T through conjugation. In fact, similar constructions show that we can perform arbitrary permutations on the elements of T via conjugation, so that there is a representation of Sn acting on the edges incident on any vertex of Γ.
5 5.1
The Lovasz Conjecture Introduction
We present a class of Cayley graphs based on the above work that the author feels are unlikely to have Hamiltonian cycles. The author has run the algorithm of Angluin and Valiant [14] repeatedly on the Cayley graph for S7 and has not yet found any Hamiltonian cycles. Although the algorithm was not designed for sparse graphs, the author still feels that this is a strike in favor of nonhamiltonicity of this graph, especially since the maximal path found (after 154000 iterations of the algorithm) has length only 2269 as compared to the necessary length of 5040 for a Hamiltonian cycle. The source code can be found online at http://www.tjhsst.edu/˜jsteinha/angluin.cpp. Furthermore,
14
Nedela and Skoviera have shown that certain attractive counterexamples (though not the ones studied here), if they exist, must occur in groups similar to Sn and An [10]. We will take the Cayley graph Γ with G = S2k−1 generated by σ = (1 2 . . . k) and τ = (2k − 1 2k − 2 . . . k), for k > 2 even. Suppose that there exists a Hamiltonian cycle C. Let x = (1 2k − 1)(2 2k − 2) . . . (k − 1 k + 1). Since the generating set is closed under conjugation by x, then xCx−1 is also a Hamiltonian cycle. It cannot be the same cycle, as the vertices of the two cycles coincide |CG (x)| times, where CG (g) denotes the centralizer of g in G (so any cyclic shift will not map (e) to (e), and one cycle can’t be the reverse of the other as otherwise they would have intersection size 1 or 2). Also note that Γ is equivalent to the graph described in Lemma 4.6 when n = 2, so that it is in fact edge-transitive (thus there must be a Hamiltonian cycle going through every edge of Γ), and we in fact have a complete characterization of its automorphism group by Lemma 4.4. Finally, note that the given Cayley graphs are always bipartite by considering even and odd permutations.
5.2
Quasi-hamiltonicity
We continue our analysis of Γ from the perspective of quasi-hamiltonicity, described in [7]. We first stop to offer a stronger version of Gutin and Yeo’s quasi-hamiltonicity for undirected graphs. We let Γ denote an undirected graph and V (Γ) and E(Γ) denote its edges and vertices, respectively. We will also let R ⊂ V (Γ). Definition A cycle factor in an undirected graph Γ is a subgraph of Γ such that every vertex has degree 2. Definition Let QH1 (Γ, R) := {e ∈ E(Γ) | e ∪ R is in a cycle factor }. For k > 1, let QHk (Γ, R) := {e ∈ E(Γ) | QHk−1 (Γ, e∪R) is connected }. Then we say that Γ is k-quasi-hamiltonian if QHk (Γ, {}) is connected. Obviously k-quasi-hamiltonicity in an undirected graph implies k-quasi-hamiltonicity in the associated digraph. Indeed, it is equivalent to k-quasi-hamiltoncity for digraphs if we disallow cycles of length 2 in the cycle factor. In particular, an undirected graph is Hamiltonian iff it is (n − 2)-quasihamiltonian, since this implies the existence of a cycle factor containing n − 2 connected vertices (so 15
the last two vertices must also be connected). We divert ourselves slightly to continue our generalization of Gutin and Yeo’s discussion from an algorithmic standpoint, which may prove useful in the future for checking the Hamiltonicity or quasi-hamiltonicity of Cayley graphs. Theorem 5.1 Given Γ, define the bipartite graph B(Γ) to have vertex set T1 = {x1 , . . . , xm } ∪ T2 = {y1 , . . . , ym }, where m = |V (Γ)|, and there exists a directed edge from xi to xj iff vertices i and j are adjacent in Γ. Create a flow network where each edge in B(Γ) has capacity 1 and there is a source s with an edge of capacity 2 into every vertex in T1 , similarly an edge of capacity 2 from every vertex in T2 into a sink t. Then there exists a cycle factor in Γ containing R iff there exists a flow of 2m from s to t, such that all edges pertaining to elements of R have flow going through them. Proof Suppose that there exists a cycle factor of Γ containing R. Push flow through xi yj and xj yi iff the edge ij is in the cycle factor. This gives the desired flow. Now suppose that we have such a flow and wish to construct a cycle factor. It is well-known that we can “force” flow to go through an edge by finding an augmenting path containing that edge and then not adding the back-flow through that edge when we push flow through the augmenting path. Thus asking for the existence of such a flow is equivalent to forcing flow through all of the edges pertaining to R (for brevity, from now on we will call this “forcing flow through R”) and asking for the existence of a flow of 2(m − |R|) in the resulting graph. Since any choice of augmenting paths must give us the same maximum flow, we can choose any set of augmenting paths that forces flow through R. In particular, given any augmenting path P, we can define another path r(P) to be the path obtained by replacing all instances of xi with yi (and vice versa) and reversing the orientation of each edge in P. Note that P and r(P) are edge-disjoint since Γ contains no self-loops. If whenever we augment by a path P, we also augment by r(P), then it will be true that P is an augmenting path iff r(P) is an augmenting path. In particular, we do this while forcing flow through R. We then continue to do this while performing the maxflow algorithm. By the symmetry of our algorithm, after we have completed it there will be flow through an edge xi yj iff there is flow through an edge xj yi . Now take the subgraph Γ0 of Γ formed by all edges ij such that there is flow through xi yj in B(Γ). Since we have a flow of 2m by assumption, every vertex in Γ0 has degree 2, thus is a cycle factor, completing the theorem.
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We furthermore extend the well-known lemma (see [7]) that all k-regular digraphs have a cycle factor. Lemma 5.2 All k-regular undirected graphs Γ have a cycle factor for even k. Proof Take an arbitrary Eulerian tour of Γ, and orient each edge in the direction it was traveled in the tour. This gives a k2 -regular digraph, which has a cycle factor because it is a regular digraph. But this contains no loops on two vertices, thus corresponds to a cycle factor in Γ, completing the lemma. It thus follows that Γ is 1-quasi-hamiltonian. Indeed, due to the high degree of symmetry of these graphs, if Γ does not contain a Hamiltonian cycle then it is (intuitively) likely to have a quasihamiltonicity number sufficiently high that it is infeasible to check. We would thus like a more efficient block to Hamiltonicity for Cayley graphs. Definition A subset T of a group G is said to have a left coset partition if there exists a set S such that s1 T and s2 T are disjoint for distinct s1 , s2 ∈ S, and such that ST = G. Definition A cycle factor is said to be symmetric if it is also a left coset partition. Note that any Hamiltonian cycle is also a symmetric cycle factor. We can thus define the analogous form of quasi-hamiltonicity where all cycle factors are required to be symmetric. Given a sufficiently crisp characterization of sets with coset partitions, it seems likely that a more effective algorithm for Hamiltonicity blocks could be designed.
6
Conclusion and Open Problems
The nature of Hamiltonian cycles in Cayley graphs remains a difficult topic. We have presented a class of “small” Cayley graphs in the sense that they are sparse with relatively large automorphism groups. There are still many aspects of these graphs to explore. As mentioned before, a characterization of left coset partitions is important for algorithmic as well as theoretical purposes. A minor but interesting detail of this paper is the dependence of our bound on X0 (A) on the existence of certain primes. There is no reason to believe that this bound should be strict, and so a 17
more complete understanding of C-trees may be reached by a more precise study of the properties of X0 (A). If the bound is given by explicit constructions, then the result of such a study would also be smaller Cayley graphs that would be more feasible to analyze empirically. Disregarding our poor understanding of X0 (A), C-trees have been effectively characterized. With this stepping stone, it would be useful to define some more concepts related to C-graphs (in a structurally interesting way). After a tree, the next simplest definition to make is that of a cycle. For a possible idea, we will borrow ideas from matroid theory. We call a set independent if it is a subset of a tree, and dependent otherwise. A simple cycle is then a subset of T that is dependent, but whose proper subsets are all independent. Of course, any other algebraic properties of graphs that could carry over to C-graphs would also be interesting. The author believes that an alternate definition of C-tree leading to a nice matroid structure on the power set of T would make all remaining generalizations transparent. The trees under this structure would also most likely lead to even more structured Cayley graphs. We have fully characterized the automorphism groups of certain C-trees. We would like a characterization, at the very least, of arbitrary split sets consisting of k-cycles. We conjecture that Theorem 4.1 holds for all such sets. Similarly, we seek a generalization of Feng’s theorem regarding the isomorphism between Aut(T ra(T )) and Aut(Sn , T ) that gives a relation between the automorphism groups of Cyc(T ) and Aut(Sn , T ).
7
Appendix
7.1
A Bound on Dirichlet’s Theorem
Theorem 7.1 For each n > 1, there is a prime of the form kn + 1 that divides Φn (n). Proof We claim that if p|Φn (j), then either p|n or p ≡ 1 (mod n). But Φn (n) ≡ 1 (mod n), so we Q must have the second case. Additionally, Φn (n) = n − ξ, for each primitive root of unity ξ. But
Y
n−ξ =
qY qY qY Y ¯ = (n − ξ)(n − ξ) n2 + 1 − 2n cos(θ) > n2 + 1 − 2n = n−1≥1
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so we must have some prime dividing Φn (n), and we are done.
7.2
Case Analysis for 4-cycles
We are essentially asking for a0 and b0 such that ab0 a0 b = (e), or equivalently bab0 a0 = (e). Thus (for the supports of ab and a0 b0 to be the same) a0 b0 ∈ {ab, ab−1 , a−1 b, a−1 b−1 , ba}. a0 cannot be b−1 and b0 cannot be a−1 since this would correspond to a path doubling back on itself. Since the product of two split k-cycles is a 2k − 1-cycle, abab is a 2k − 1-cycle, and in particular not the identity. abab−1 = a(bab−1 ) is the product of two k-cycles with different supports, and so is again not the identity. Similar logic holds for aba−1 b = (aba−1 )b. aba−1 b−1 = (e) implies that ab = ba, which is what we want. Finally, abba = (e) implies aabb = (e), which is impossible since aa and bb have different supports.
7.3
Case Analysis for Commutators
The full details of this argument can be found at http://www.tjhsst.edu/˜jsteinha/ Cayley.pdf. The crux of the argument is repeated use of symmetry, which eventually shows that all interesting cases WLOG start with abab−1 . We then list out all products of four generators such that no two consecutive generators commute or represent the same element. By looking at what compositions of permutations can send 1 back to 1, we reduce essentially to 6 remaining cases, which are easy to check through simple calculations. This completes our case analysis.
7.4
Induction Argument for Lemma 4.4
We have already shown that φ(a)φ(b) = φ(ab) for any automorphism φ of Γ fixing (e). We have the following lemma: Lemma 7.2 If φ is a (graph) automorphism of Γ, then so is φy = φ(y −1 )φ(yx). The proof is a routine verification. Now, we wish to show by induction that
φ(t1 t2 . . . tn ) = φ(t1 )φ(t2 ) . . . φ(tn ) 19
for all φ ∈ Γ. Now note that
φ(t1 t2 . . . tn ) = φ(t1 )φt1 (t2 . . . tn ) = φ(t1 )φt1 (t2 ) . . . φt1 (tn ) = φ(t1 ) . . . φ(tn ) where the equality between the second and third expressions follows by the inductive step. This completes our induction.
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